Oh No. Did We Fail Both?

Probability Level 1

Martin's class has 40 students who took an examination in History and Mathematics. The teacher announced that 28 of them passed in Mathematics, 29 of them passed in History, and 20 of them passed in both subjects.

How many students failed both subjects?

12 20 6 3

17 solutions

Victor Paes Plinio
May 24, 2014

I used the Venn's Diagram.

First the intersection is 20.

In history is $29-20=9$

Mathematics is $28-20=8$

Make this sum: $9+8+20=37$

Subtracting the success studentas from the total students we get the failed students.

$40-37=3$

The answer is $\boxed { 3}$

It is very clear that students passing in both subjects will also be counted in students passing in individual subjects. So we subtract 20 from both figures...

Students passing only in history is 29 - 20 = 9 Students passing only in mathematics is 28 -20 = 8 Total passing students in subjects whatsoever = 9 + 8 + 20 = 37 Therefore failures = 40 - 37 = 3

Nishant Prabhu - 7 years ago

p(MuH)= P(M) + P(H) - P(M^H)=28/40 + 29/40 - 20/40 = 3/40 so p(M ^H ) = 1 -P(M U H) = 3/40 SO 3

aliki patsalidou - 7 years ago

OUT OF 40 STUDENT 20 PASSED IN BOTH SUBJECT

OF THE REMAINING 20 STUDENTS (8+9=17) PASSED IN ONE SUBJECT

THEREFORE NO. OF REMAINING STUDENTS FAILED IN BOTH SUBJECTS= 40-20-8-9=3

Zakaria Ameen - 7 years ago

hmmmm good

ali haider - 7 years ago

ahh ahh naman..

Ord Sobremonte Jr. - 7 years ago

I have also used Venn's diagram .

Chinmoyranjan Giri - 7 years ago

Joseph Lim
May 24, 2014

$\xi =40\\ Students\quad who\quad pass\quad mathematics,\quad A=28\\ Students\quad who\quad pass\quad history,\quad B=29\\ Students\quad who\quad pass\quad both\quad subjects,\quad A\cap B=20\\ \\ Find\quad students\quad who\quad failed\quad both\quad subjects,\quad A\prime \cap B\prime \\ \\ Using\quad De\quad Morgan's\quad Law,\\ A\prime \cap B\prime \\ =(A\cup B)\prime \\ =(A+B-A\cap B)\prime \\ =(28+29-20)\prime \\ =(37)\prime \\ =3$

veryyyyyyyyyyyyy good answer

Rozi Khan - 7 years ago

tactic question but simple

ayoade habibullah - 7 years ago

excellent answer!

Gita Shinde - 5 years, 7 months ago

Deepthi Prakash
May 24, 2014

first subtract 29 by 20 to get 9 plus 28 is 37. 40 minus 37 is 3.

so,3=answer correct

this how I solved it....

deepthi prakash - 7 years ago

great oooooo

Juang Bhakti Hastyadi - 7 years ago

=(29-20)+(28-20) =9+8 =17 Now, 40-(17+20) =3 Ans

joni saha - 7 years ago

Abc Def
May 24, 2014

n(A||B)=n(A) + n(B) - n(A&B)

|| Represents union and & represents intersection .

Don't know why the semicolon appears :/

Sadasiva Panicker
Oct 30, 2015

Number of students failed both subjects = 40 - (20 + 8 + 9) = 40 - 37 = 3

Meeca Taladtad
Oct 30, 2015

28+29+20=77 77-40=37 40-37=3 That's my solution.....

Michael Friese
Oct 29, 2015

20 of them passed both subjects, so we had 20 of them that failed at least one subject. 12 failures in math, 11 failures in history. total failures is 23. So, 3 must have failed both.

Wissam Akil
Jun 7, 2014

[n(aUb)’=∑-n(aUb)] ,,Where ∑=40 ,,And n(aUb)=8+9+20=37 ,,So , n(aUb)’ = 40 – 37 = 3

Ray Cuesta
Jun 6, 2014

45=(29-20)+(28-20)+20+x, x=3

Krishna Garg
Jun 4, 2014

Out of 40 students 12 failed in Maths ,and 11 failed in Histoty,total failed are 20 .So total failed in Maths and history are 12 +11 =23 bou actual failed are 20 .Therefore both failed in History and Maths are 23 -20 =3 Ans. K.K.GARG,India

Trishit Chandra
May 31, 2014

With the help of Venn's diagram We can write that, P(AUB)=P(A)+P(B)-P(A intersection B) or, P(AUB)=28+29-20 or, P(AUB)=37 And the total no. of student is 40. Then the total no. of student who failed is (40-37=3)

THE ANSWER IS 3.